What are the solutions to the equation x(5x-3)=1 ?

The solutions to the equation x(5x-3)=1 are x=(3+sqrt(29))/(10),x=(3-sqrt(29))/(10)

Find the zeros of x(5x-3)=1

The zeros of x(5x-3)=1 are x=(3+sqrt(29))/(10),x=(3-sqrt(29))/(10)

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x (5 x - 3) = 1

x = \frac{3 + 29}{10}, x = \frac{3 - 29}{10}

Decimal Notation

x = 0.83851 \dots, x = - 0.23851 \dots

Solution steps

x (5 x - 3) = 1

Expand

x (5 x - 3) : 5 x^{2} - 3 x

5 x^{2} - 3 x = 1

Move

1

to the left side

5 x^{2} - 3 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 1) = 29

x_{1, 2} = \frac{- ( - 3 ) \pm 29}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 29}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 29}{2 \cdot 5}

x = \frac{- ( - 3 ) + 29}{2 \cdot 5} : \frac{3 + 29}{10}

x = \frac{- ( - 3 ) - 29}{2 \cdot 5} : \frac{3 - 29}{10}

The solutions to the quadratic equation are:

x = \frac{3 + 29}{10}, x = \frac{3 - 29}{10}

3 x (2 x + 3) = 5 x + 6

x^{2} - 36 = 9 x

- 2 x^{2} + 4 x + 5 = 0

d^{2} = 4

(x + 2)^{3} = x^{3} + 8

What are the solutions to the equation x(5x-3)=1 ?
The solutions to the equation x(5x-3)=1 are x=(3+sqrt(29))/(10),x=(3-sqrt(29))/(10)
Find the zeros of x(5x-3)=1
The zeros of x(5x-3)=1 are x=(3+sqrt(29))/(10),x=(3-sqrt(29))/(10)